 Solution

For either scalar equation, the complementary solution isso initial velocity transients decay when $t\gg (ϵ\Omega {)}^{-1}$. In the exact corrected forced dust epicycle with aerodynamic drag, $v_{x,p}^{\prime }=u\cos \theta$ is a particular solution, and the original radial equation then gives $v_{y,p}^{\prime }=u\sin \theta /2$. ConsequentlyThis is resonant dust entrainment by an epicyclic gas wave: in the adopted long-wavelength gas approximation, the forcing is a free epicycle of the dust too, so matching the gas makes the drag zero. The convergence holds for every $ϵ>0$ in this linear model; at $ϵ=0$ transients never decay, so the zero-drag and late-time limits do not commute.

If instead the printed scalar equation is treated literally for finite $ϵ$, its particular solution isIt has amplitude $|u|/\sqrt{1+{ϵ}^2/4}$ and phase offset $\arctan (ϵ/2)$, not exact phase/amplitude agreement. Agreement in that truncated equation means only leading order as $ϵ\to 0$. This distinguishes the intended approximation from a false exact reading of the printed assertion.

There is a further order-of-limits condition: a real finite-wavelength gas wave has $\omega -\Omega \ne 0$. Leading resonant entrainment over the damping time requires $|\omega -\Omega |\ll ϵ\Omega$, or $(k{c}_s/\Omega {)}^2\ll ϵ$. Otherwise solve the dust transfer equations at the actual $\omega$; detuning can reduce the dust response even though $|k|c_s/\Omega \ll 1$.